Properties of Ellipses (Edexcel A Level Further Maths: Further Pure 1): Revision Note

You are given the Cartesian and parametric equations of an ellipse in the formulae booklet.

You are given the eccentricity formula, foci and directrices of an ellipse in the formulae booklet.

You are not given any formulae for the $a<b$ case, but you can work them out by swapping 'horizontal to vertical' and ' $a$ to $b$'.

An ellipse has the equation $\frac{x^2}{25}+\frac{y^2}{16}=1$.

(a) Calculate the coordinates of the foci.

(b) Calculate the equations of the directrices.

Answer:

(a)

Find $a$ and $b$ by comparing to the general equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\begin{array}{l}a=5\\ b=4\end{array}$

Check that $a>b$

$5>4$

Rearrange the relationship $b^2=a^2\left(1-e^2\right)$ to find $e$

• and check that $0\le e<1$

$\begin{array}{rcl}{4}^2& =& 5^2\left(1-e^2\right)\\ \frac{16}{25}& =& 1-e^2\\ e^2& =& 1-\frac{16}{25}\\ e^2& =& \frac{9}{25}\\ e& =& \frac{3}{5}\end{array}$

Calculate the foci using $\left(\pm ae,0\right)$

$\left(\pm 5\times \frac{3}{5},0\right)$

The foci have coordinates $\left(\pm 3,0\right)$

(b)

Calculate the equations of the directrices using $x=\pm \frac{a}{e}$

$x=\pm \frac{5}{\left({\displaystyle \frac{3}{5}}\right)}$

The directrices have equations $x=\pm \frac{25}{3}$

You are not given the focus-directrix property in the exam (you must learn it).

An ellipse with foci $F$ and $F\text{'}$ and directrices $x=\pm \frac{a}{e}$ is shown below.

The point $P$ on the ellipse has coordinates $\left(x,y\right)$ and the points $D$ and $D\text{'}$ are on the directrices, at the same height as $P$.

Using only the focus-directrix property,

(a) prove that $PF\text{'}+PF=2a$

(b) derive the Cartesian equation of an ellipse, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $b^2=a^2\left(1-e^2\right)$

Answer:

(a)

Use the focus-directrix property on $P$, $F\text{'}$ and $D\text{'}$ and again on $P$, $F$ and $D$

$\frac{PF\text{'}}{PD\text{'}}=e\mathrm{and}\frac{PF}{PD}=e$

Rearrange to make $PF\text{'}$ and $PF$ the subjects

$PF\text{'}=ePD\text{'}\mathrm{and}PF=ePD$

Add together $PF\text{'}$ and $PF$

$PF\text{'}+PF=ePD\text{'}+ePD$

Factorise out $e$

$PF\text{'}+PF=e\left(PD\text{'}+PD\right)$

Use that $PD\text{'}+PD$ is the total distance between the two directrices $x=\pm \frac{a}{e}$

$PD\text{'}+PD=\frac{a}{e}-\left(-\frac{a}{e}\right)=\frac{2a}{e}$

Substitute this back into $PF\text{'}+PF$

$PF\text{'}+PF=e\left(\frac{2a}{e}\right)$

Simplify

$PF\text{'}+PF=2a$

(b)

Use the focus-directrix property on $P$, $F$ and $D$ (draw on lines $PF$ and $PD$)

$\frac{PF}{PD}=e$

It helps to draw lengths $x$ and $y$ from $P\left(x,y\right)$ on the diagram and the foci $\left(\pm ae,0\right)$

Create a right-angled triangle whose hypotenuse is $PF$ with base $\left(ae-x\right)$ and height $y$

Use Pythagoras' theorem to find $P{F}^2$

$P{F}^2={\left(ae-x\right)}^2+y^2$

Find the length $PD$ from $x$ to the directrix

$PD=\left(\frac{a}{e}-x\right)$

Rearrange $\frac{PF}{PD}=e$ to make $P{F}^2$ the subject

$P{F}^2=e^{2}P{D}^2$

Substitute in expressions for $P{F}^2$ and $P{D}^2$ from above

${\left(ae-x\right)}^2+y^2=e^2{\left(\frac{a}{e}-x\right)}^2$

Expand, cancel and factorise

$\begin{array}{rcl}{a}^2{e}^2-2aex+x^2+y^2& =& e^2\left(\frac{a^2}{e^2}-\frac{2ax}{e}+x^2\right)\\ a^2{e}^2\cancel{-2aex}+x^2+y^2& =& a^2\cancel{-2aex}+e^2{x}^2\\ \left(1-e^2\right)x^2+y^2& =& a^2\left(1-e^2\right)\end{array}$

Divide both sides by $a^2\left(1-e^2\right)$

$\frac{x^2}{a^2}+\frac{y^2}{a^2\left(1-e^2\right)}=1$

This is now in the correct form of an ellipse

The Cartesian equation is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $b^2=a^2\left(1-e^2\right)$